实例一、对鸢尾花使用SVM进行分类

#1、加载数据
data(iris)
#2、创建测试集和训练集数据
index <- sample(1:2, nrow(iris), prob=c(0.8, 0.2), replace=T)
train_iris <- iris[index==1, ]
test_iris <- iris[index==2, ]
#3、建模
library(e1071)
model_iris <- svm(Species~., data=train_iris, type="C-classification", cost=10, kernal="radial", gamma=0.1, scale=F)
#4、模型评估
model_iris

##
## Call:
## svm(formula = Species ~ ., data = train_iris, type = "C-classification",
##
cost = 10, kernal = "radial", gamma = 0.1, scale = F)
##
##
## Parameters:
##
SVM-Type:
C-classification
##
SVM-Kernel:
radial
##
cost:
10
##
gamma:
0.1
##
## Number of Support Vectors:
26

pred <- predict(model_iris, train_iris)
mean(pred==train_iris[, 5])

## [1] 0.983871

table(pred, train_iris[, 5])

##
## pred
setosa versicolor virginica
##
setosa
44
0
0
##
versicolor
0
39
0
##
virginica
0
2
39

#5、预测
pred_iris <- predict(model_iris, test_iris)
mean(pred_iris==test_iris[, 5])

## [1] 1

table(pred_iris, test_iris[, 5])

##
## pred_iris
setosa versicolor virginica
##
setosa
6
0
0
##
versicolor
0
9
0
##
virginica
0
0
11

#6、修改cost值来查看新的结果，默认为10-fold CV
model_iris1 <- svm(Species~., train_iris, kernal="radial", cost=0.1, scale = F)
pred1 <- predict(model_iris1, test_iris)
mean(pred1==test_iris[, 5])

## [1] 0.9615385

table(pred1, test_iris[, 5])

##
## pred1
setosa versicolor virginica
##
setosa
6
0
0
##
versicolor
0
9
1
##
virginica
0
0
10

#7、使用tune()函数调整cost的值，默认为10-fold CV
model_tune <- tune(svm, Species~., data=train_iris, kernal="radial", ranges=list(cost=c(0.001, 0.01, 0.1, 1, 5, 10, 100)))
summary(model_tune)

##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
##
cost
##
1
##
## - best performance: 0.04871795
##
## - Detailed performance results:
##
cost
error dispersion
## 1 1e-03 0.73205128 0.11830739
## 2 1e-02 0.73205128 0.11830739
## 3 1e-01 0.11282051 0.12390215
## 4 1e+00 0.04871795 0.05751000
## 5 5e+00 0.05705128 0.05567459
## 6 1e+01 0.05705128 0.06813861
## 7 1e+02 0.05641026 0.06619105

str(model_tune)

## List of 8
##
$ best.parameters :'data.frame':
1 obs. of
1 variable:
##
..$ cost: num 1
##
..- attr(*, "out.attrs")=List of 2
##
.. ..$ dim
: Named int 7
##
.. .. ..- attr(*, "names")= chr "cost"
##
.. ..$ dimnames:List of 1
##
.. .. ..$ cost: chr [1:7] "cost=1e-03" "cost=1e-02" "cost=1e-01" "cost=1e+00" ...
##
$ best.performance: num 0.0487
##
$ method
: chr "svm"
##
$ nparcomb
: int 7
##
$ train.ind
:List of 10
##
..$ (0.877,13.3]: int [1:111] 103 113 64 9 26 74 90 20 85 78 ...
##
..$ (13.3,25.6] : int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (25.6,37.9] : int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (37.9,50.2] : int [1:111] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (50.2,62.5] : int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (62.5,74.8] : int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (74.8,87.1] : int [1:111] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (87.1,99.4] : int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (99.4,112]
: int [1:112] 111 66 91 15 14 63 46 124 119 12 ...
##
..$ (112,124]
: int [1:111] 111 66 91 15 14 63 46 124 119 12 ...
##
..- attr(*, "dim")= int 10
##
..- attr(*, "dimnames")=List of 1
##
.. ..$ : chr [1:10] "(0.877,13.3]" "(13.3,25.6]" "(25.6,37.9]" "(37.9,50.2]" ...
##
$ sampling
: chr "10-fold cross validation"
##
$ performances
:'data.frame':
7 obs. of
3 variables:
##
..$ cost
: num [1:7] 1e-03 1e-02 1e-01 1e+00 5e+00 1e+01 1e+02
##
..$ error
: num [1:7] 0.7321 0.7321 0.1128 0.0487 0.0571 ...
##
..$ dispersion: num [1:7] 0.1183 0.1183 0.1239 0.0575 0.0557 ...
##
$ best.model
:List of 30
##
..$ call
: language best.tune(method = svm, train.x = Species ~ ., data = train_iris,
ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100)), kernal = "radial")
##
..$ type
: num 0
##
..$ kernel
: num 2
##
..$ cost
: num 1
##
..$ degree
: num 3
##
..$ gamma
: num 0.25
##
..$ coef0
: num 0
##
..$ nu
: num 0.5
##
..$ epsilon
: num 0.1
##
..$ sparse
: logi FALSE
##
..$ scaled
: logi [1:4] TRUE TRUE TRUE TRUE
##
..$ x.scale
:List of 2
##
.. ..$ scaled:center: Named num [1:4] 5.8 3.06 3.66 1.15
##
.. .. ..- attr(*, "names")= chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##
.. ..$ scaled:scale : Named num [1:4] 0.82 0.437 1.751 0.755
##
.. .. ..- attr(*, "names")= chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##
..$ y.scale
: NULL
##
..$ nclasses
: int 3
##
..$ levels
: chr [1:3] "setosa" "versicolor" "virginica"
##
..$ tot.nSV
: int 46
##
..$ nSV
: int [1:3] 8 19 19
##
..$ labels
: int [1:3] 1 2 3
##
..$ SV
: num [1:46, 1:4] -1.709 -0.124 -0.49 -0.855 -0.977 ...
##
.. ..- attr(*, "dimnames")=List of 2
##
.. .. ..$ : chr [1:46] "9" "16" "21" "24" ...
##
.. .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##
..$ index
: int [1:46] 9 15 19 21 23 29 37 38 45 46 ...
##
..$ rho
: num [1:3] -0.0606 0.0842 0.0634
##
..$ compprob
: logi FALSE
##
..$ probA
: NULL
##
..$ probB
: NULL
##
..$ sigma
: NULL
##
..$ coefs
: num [1:46, 1:2] 0.0895 0.8159 0 0.6514 0.6058 ...
##
..$ na.action
: NULL
##
..$ fitted
: Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
##
.. ..- attr(*, "names")= chr [1:124] "1" "2" "3" "4" ...
##
..$ decision.values: num [1:124, 1:3] 1.19 1.06 1.17 1.1 1.17 ...
##
.. ..- attr(*, "dimnames")=List of 2
##
.. .. ..$ : chr [1:124] "1" "2" "3" "4" ...
##
.. .. ..$ : chr [1:3] "setosa/versicolor" "setosa/virginica" "versicolor/virginica"
##
..$ terms
:Classes 'terms', 'formula' length 3 Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width
##
.. .. ..- attr(*, "variables")= language list(Species, Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)
##
.. .. ..- attr(*, "factors")= int [1:5, 1:4] 0 1 0 0 0 0 0 1 0 0 ...
##
.. .. .. ..- attr(*, "dimnames")=List of 2
##
.. .. .. .. ..$ : chr [1:5] "Species" "Sepal.Length" "Sepal.Width" "Petal.Length" ...
##
.. .. .. .. ..$ : chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##
.. .. ..- attr(*, "term.labels")= chr [1:4] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width"
##
.. .. ..- attr(*, "order")= int [1:4] 1 1 1 1
##
.. .. ..- attr(*, "intercept")= num 0
##
.. .. ..- attr(*, "response")= int 1
##
.. .. ..- attr(*, ".Environment")=<environment: R_GlobalEnv>
##
.. .. ..- attr(*, "predvars")= language list(Species, Sepal.Length, Sepal.Width, Petal.Length, Petal.Width)
##
.. .. ..- attr(*, "dataClasses")= Named chr [1:5] "factor" "numeric" "numeric" "numeric" ...
##
.. .. .. ..- attr(*, "names")= chr [1:5] "Species" "Sepal.Length" "Sepal.Width" "Petal.Length" ...
##
..- attr(*, "class")= chr [1:2] "svm.formula" "svm"
##
- attr(*, "class")= chr "tune"

#获取最好的模型
model_best <- model_tune$best.model
summary(model_best)

##
## Call:
## best.tune(method = svm, train.x = Species ~ ., data = train_iris,
##
ranges = list(cost = c(0.001, 0.01, 0.1, 1, 5, 10, 100)),
##
kernal = "radial")
##
##
## Parameters:
##
SVM-Type:
C-classification
##
SVM-Kernel:
radial
##
cost:
1
##
gamma:
0.25
##
## Number of Support Vectors:
46
##
##
( 8 19 19 )
##
##
## Number of Classes:
3
##
## Levels:
##
setosa versicolor virginica

#预测
pred_tune <- predict(model_best, test_iris)
mean(pred_tune==test_iris[, 5])

## [1] 1

table(pred_tune, test_iris[, 5])

##
## pred_tune
setosa versicolor virginica
##
setosa
6
0
0
##
versicolor
0
9
0
##
virginica
0
0
11

实例二：寻找最优参数

使用支持向量机实现二元分类器，使用的数据是来自MASS包的cats数据集。在本例中你将尝试使用体重和心脏重量来预测一只猫的性别。

#1、加载数据
data(cats, package="MASS")
str(cats)

## 'data.frame':
144 obs. of
3 variables:
##
$ Sex: Factor w/ 2 levels "F","M": 1 1 1 1 1 1 1 1 1 1 ...
##
$ Bwt: num
2 2 2 2.1 2.1 2.1 2.1 2.1 2.1 2.1 ...
##
$ Hwt: num
7 7.4 9.5 7.2 7.3 7.6 8.1 8.2 8.3 8.5 ...

summary(cats)

##
Sex
Bwt
Hwt
##
F:47
Min.
:2.000
Min.
: 6.30
##
M:97
1st Qu.:2.300
1st Qu.: 8.95
##
Median :2.700
Median :10.10
##
Mean
:2.724
Mean
:10.63
##
3rd Qu.:3.025
3rd Qu.:12.12
##
Max.
:3.900
Max.
:20.50

#2、创建训练集和测试集数据
index <- sample(1:2, nrow(cats), prob=c(0.7, 0.3), replace=T)
train_cats <- cats[index==1, ]
test_cats <- cats[index==2, ]
#3、建模
library(e1071)
#1）线性核函数linear SVM
model_linear <- svm(Sex~., train_cats, kernal="linear", cost=10, scale = F)
#预测
pred <- predict(model_linear, test_cats[])
mean(pred==test_cats$Sex)

## [1] 0.7560976

table(pred, test_cats$Sex)

##
## pred
F
M
##
F
7
6
##
M
4 24

print(model_linear)

##
## Call:
## svm(formula = Sex ~ ., data = train_cats, kernal = "linear",
##
cost = 10, scale = F)
##
##
## Parameters:
##
SVM-Type:
C-classification
##
SVM-Kernel:
radial
##
cost:
10
##
gamma:
0.5
##
## Number of Support Vectors:
55

#2)创建径向支持向量机：radial SVM
model_radial <- svm(Sex~., train_cats, kernal="radial", cost=10, scale=F)
#预测
pred <- predict(model_radial, test_cats)
mean(pred==test_cats$Sex)

## [1] 0.7560976

table(pred, test_cats$Sex)

##
## pred
F
M
##
F
7
6
##
M
4 24

print(model_radial)

##
## Call:
## svm(formula = Sex ~ ., data = train_cats, kernal = "radial",
##
cost = 10, scale = F)
##
##
## Parameters:
##
SVM-Type:
C-classification
##
SVM-Kernel:
radial
##
cost:
10
##
gamma:
0.5
##
## Number of Support Vectors:
55

#4、寻找最优参数：可以使用tune.svm()函数，来寻找svm()函数的最优参数。
model_tuned <- tune.svm(Sex~., data=train_cats, gamma=10^(-6:-1),cost=10^(1:2))
summary(model_tuned)

##
## Parameter tuning of 'svm':
##
## - sampling method: 10-fold cross validation
##
## - best parameters:
##
gamma cost
##
0.001
100
##
## - best performance: 0.2172727
##
## - Detailed performance results:
##
gamma cost
error dispersion
## 1
1e-06
10 0.3463636
0.2105330
## 2
1e-05
10 0.3463636
0.2105330
## 3
1e-04
10 0.3463636
0.2105330
## 4
1e-03
10 0.3681818
0.2027474
## 5
1e-02
10 0.2272727
0.1177537
## 6
1e-01
10 0.2363636
0.1363973
## 7
1e-06
100 0.3463636
0.2105330
## 8
1e-05
100 0.3463636
0.2105330
## 9
1e-04
100 0.3681818
0.2027474
## 10 1e-03
100 0.2172727
0.1243865
## 11 1e-02
100 0.2272727
0.1432920
## 12 1e-01
100 0.2354545
0.1426143

#结果证明，当cost为10，gamma为0.1时产生最小的错误率。利用这些参数训练径向支持向量机。
#5、利用调整后的最优参数：cost=10, gamma=0.1重新建模
model_cats <- svm(Sex~., train_cats, kernal="radial", cost=10, gamma=0.1, scale = F)
#6、预测
pred <- predict(model_cats, test_cats)
mean(pred==test_cats$Sex)

## [1] 0.7560976

table(pred, test_cats$Sex)

##
## pred
F
M
##
F
8
7
##
M
3 23

实例三：kernlab包中的ksvm()函数实现支持向量机

#1、加载数据
data(iris)
#2、创建训练集和测试集数据
index <- sample(1:2, nrow(iris), prob=c(0.7, 0.3), replace = T)
train_iris <- iris[index==1, ]
test_iris <- iris[index==2, ]
#3、建模
library(kernlab)
model <- ksvm(Species~., train_iris, kernal="rbfdot", type="C-bsvc", kpar=list(sigma=0.1), C=10, prob.model=T)
#4、预测
pred <- predict(model, test_iris)
mean(pred==test_iris[, 5])

## [1] 0.9642857

table(pred, test_iris[, 5])

##
## pred
setosa versicolor virginica
##
setosa
18
0
0
##
versicolor
0
19
0
##
virginica
0
2
17